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ivann1987 [24]
2 years ago
15

If 25 ml of 1.50M LiOH are neutralized by 15ml of nitric acid what is the molarity of HNO3

Chemistry
2 answers:
Jobisdone [24]2 years ago
8 0
LiOH + HNO3 --> H2O + LiNO3
OH- + H+  --> H2O
OH - = 1.5M * .025L = .0375 mol
.0375mol/.04L = 0.9375M
Since it's at the equivalence point, OH- = H+ concentration

[HNO3] = 0.9375M
nasty-shy [4]2 years ago
8 0

Answer:

Ca = 2.5M

Explanation:

Volume of Acid (Va) = 15ml

Concentration of acid (Ca) = ?

Volume of base (Vb) = 25

Concentration of base (Cb) = 1.50M

The balanced chemical equation is given as;

LiOH + HNO3 --> LiNO3 + H2O

The formular relaing the parameters is given as;

(CaVa) / (CbVb) = Na/Nb

Where Na = Number of moles of Acid = 1

Nb = Number of moles of Base = 1

Making Ca subject of interest;

Ca = (Na * Cb * Vb) / (Nb * Va)

Ca = (1 * 1.50 * 25) / (1 * 15)

Ca = 37.5 / 15

Ca = 2.5M

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How would you prepare 500 mL of 0.360 M solution of CaCl2 from<br> solid CaCl2?
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We need to measure 20.0 grams of CaCl₂ to prepare 500 mL of 0.360 M solution.

First, we need to determine the required moles of CaCl₂. We have 500 mL (0.500 L) of a 0.360 M solution (0.360 moles of CaCl₂ per liter of solution).

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Then, we will convert 0.180 moles to grams using the molar mass of CaCl₂ (110.98 g/mol).

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To prepare the solution, we weigh 20.0 g of CaCl₂ and add it to a beaker with enough distilled water to dissolve it. We stir it, heat it if necessary, and when we have a solution, we transfer it to a 500 mL flask and complete it to the mark with distilled water.

We need to measure 20.0 grams of CaCl₂ to prepare 500 mL of 0.360 M solution.

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